Why does London equation not hold for normal conductors? In exercise 13.2 from Fetter, Walecka - Quantum Theory of Many-Particles Systems it is asked to obtain London equation as the Euler-Lagrange equation for the Helmholtz free energy and to discuss its validity. Free energy for arbitrary field $\vec{B}$ is:
$$F_s(T,B) = F_s(T,0) + \int d^3x \left[ \frac{1}{8\pi} B^2 + \frac{1}{2}n_smv^2\right]$$
Where $B$ is the magnetic field, $n_s$ is the density of superconducting charge carriers and $m$ is their mass.
Expressing $\vec{v}$ with respect to $\vec{B}$ ( $\frac{4\pi}{c} \vec{j} = \operatorname{curl} \vec{B} $)  and imposing $$ \frac{\delta}{\delta B_i}F_s(T,B) =0 $$ one straightforwardly obtains "London" equation
 $$\nabla^2 \vec{B} = \frac{1}{\lambda^2}\vec{B}.$$
The problem is: why should this not hold for a normal conductor? 
 A: It actually does - sort of.
Remember that the free energy functional is a functional of two fields, the superconducting wavefunction $\psi$ and the magnetic field $\mathbf B$ (or the vector potential $\mathbf A$, if you prefer).  The equilibrium state is achieved when the functional is minimized with respect to both $\psi$ and $\mathbf B$.
As a result, you end up with two coupled differential equations which must be solved together.  You've written one of them, a differential equation for $\mathbf B$, but in the process you have made certain assumptions about $\psi$ (or really, $|\psi|^2$) which may not be consistent with the solution to the other.
I would suggest that you consider a superconducting metal and then imagine what about the above equations (and their solutions) changes as $T$ increases past $T_c$.  In particular, don't neglect what happens to $F_s(T,0)$, and what influence that has on $\psi$.

More explicitly, variation of $F_s$ with respect to $\psi$ or $n_s$ yields $|\psi|^2=n_s=0$ for $T>T_c$.  This being the case, the problem reduces to the minimization of 
$$\int d^3x \frac{B^2}{8\pi}$$
subject to any prescribed boundary conditions.  However, setting the functional derivative to zero does not produce a differential equation (it simply demands that $\mathbf B=0$), reflecting the fact that the inclusion of boundary conditions generically requires a differential equation to accommodate them.  As it stands, prescribing $\mathbf B$ on the boundary and that $\frac{\delta F_s}{\delta \mathbf B}\sim \mathbf B = 0$ on leaves the system overdetermined and inconsistent.
A: The reason is the $\frac{1}{2}n_s m v^2$ term. In superconductors, the kinetic energy can form a significant fraction of an electron's (or, really, a Cooper pair's) energy, while in ordinary conductors the contribution of the kinetic energy is negligible. This is because in a superconductor the pairs move together in a rigid, coherent state over the entire macroscopic volume of the sample, while in a normal metal the electrons' wave functions are incoherent beyond a very short microscopic length scale.
